140. what is the ionic form of the following unbalanced equation?
$\text{mno}_2 + \text{hno}_2 \
ightarrow \text{mn}(\text{no}_3)_2 + \text{h}_2\text{o}$
a. $\text{mno}_2 + \text{hno}_2 \
ightarrow \text{mn}^{2+} + \text{no}_3^- + \text{h}_2\text{o}$
b. $\text{mno}_2 + \text{h}^+ + \text{no}_2^- \
ightarrow \text{mnno}_3 + \text{h}_2\text{o}$
c. $\text{mno}_2 + \text{h}^+ + \text{no}_2^- \
ightarrow \text{mn}^{2+} + \text{no}_3^- + \text{h}_2\text{o}$
d. $\text{mn}^{2+} + \text{o}^{2-} + \text{h}^+ + \text{no}_2^- \
ightarrow \text{mn}^{2+} + \text{no}_3^- + \text{h}_2\text{o}$

141. what is a balanced equation for the redox reaction represented by the two half-reactions below?
$\text{br}_2 + 2\text{e}^- \
ightarrow 2\text{br}^-$
$\text{na} \
ightarrow \text{na}^+ + \text{e}^-$
a. $\text{na} + \text{br}_2 + \text{e}^- \
ightarrow \text{na}^+ + 2\text{br}^-$
b. $\text{na} + \text{br}_2 \
ightarrow \text{na}^+ + 2\text{br}^-$
c. $2\text{na} + \text{br}_2 \
ightarrow 2\text{na}^+ + 2\text{br}^-$
d. $\text{na} + \text{br}_2 + 2\text{e}^- \
ightarrow \text{na}^+ + 2\text{br}^- + \text{e}^-$

142. which metal is the most easily oxidized?
a. highly active metal
b. moderately active metal
c. slightly active metal
d. an inactive metal

143. which of these metals ions are most easily reduced?
a. iron
b. mercury
c. aluminum
d. potassium

144. which metal will plate a lead strip that is dipped into a solution of its ions?
a. potassium
b. calcium
c. mercury
d. cadmium

145. how is the cell potential of a voltaic cell calculated?
a. $e_{\text{cell}} = e_{\text{red}} - e_{\text{oxid}}$
b. $e_{\text{cell}} = e_{\text{oxid}} - e_{\text{red}}$
c. $e_{\text{cell}} = e_{\text{red}} + e_{\text{oxid}}$
d. $e_{\text{cell}} = e_{\text{red}} + e_{\text{oxid}}$

146. how can a redox reaction be used as a source of electrical energy?
a. two half-reactions must be physically separated.
b. one half-reaction must involve two metals.
c. two half-reactions must involve more than one electron.
d. one half-reaction must use a metal wire electrode.

Question

140. what is the ionic form of the following unbalanced equation?
$\text{mno}_2 + \text{hno}_2 \
ightarrow \text{mn}(\text{no}_3)_2 + \text{h}_2\text{o}$
a. $\text{mno}_2 + \text{hno}_2 \
ightarrow \text{mn}^{2+} + \text{no}_3^- + \text{h}_2\text{o}$
b. $\text{mno}_2 + \text{h}^+ + \text{no}_2^- \
ightarrow \text{mnno}_3 + \text{h}_2\text{o}$
c. $\text{mno}_2 + \text{h}^+ + \text{no}_2^- \
ightarrow \text{mn}^{2+} + \text{no}_3^- + \text{h}_2\text{o}$
d. $\text{mn}^{2+} + \text{o}^{2-} + \text{h}^+ + \text{no}_2^- \
ightarrow \text{mn}^{2+} + \text{no}_3^- + \text{h}_2\text{o}$

141. what is a balanced equation for the redox reaction represented by the two half-reactions below?
$\text{br}_2 + 2\text{e}^- \
ightarrow 2\text{br}^-$
$\text{na} \
ightarrow \text{na}^+ + \text{e}^-$
a. $\text{na} + \text{br}_2 + \text{e}^- \
ightarrow \text{na}^+ + 2\text{br}^-$
b. $\text{na} + \text{br}_2 \
ightarrow \text{na}^+ + 2\text{br}^-$
c. $2\text{na} + \text{br}_2 \
ightarrow 2\text{na}^+ + 2\text{br}^-$
d. $\text{na} + \text{br}_2 + 2\text{e}^- \
ightarrow \text{na}^+ + 2\text{br}^- + \text{e}^-$

142. which metal is the most easily oxidized?
a. highly active metal
b. moderately active metal
c. slightly active metal
d. an inactive metal

143. which of these metals ions are most easily reduced?
a. iron
b. mercury
c. aluminum
d. potassium

144. which metal will plate a lead strip that is dipped into a solution of its ions?
a. potassium
b. calcium
c. mercury
d. cadmium

145. how is the cell potential of a voltaic cell calculated?
a. $e_{\text{cell}} = e_{\text{red}} - e_{\text{oxid}}$
b. $e_{\text{cell}} = e_{\text{oxid}} - e_{\text{red}}$
c. $e_{\text{cell}} = e_{\text{red}} + e_{\text{oxid}}$
d. $e_{\text{cell}} = e_{\text{red}} + e_{\text{oxid}}$

146. how can a redox reaction be used as a source of electrical energy?
a. two half-reactions must be physically separated.
b. one half-reaction must involve two metals.
c. two half-reactions must involve more than one electron.
d. one half-reaction must use a metal wire electrode.

Accepted Answer

{ "quality": "clear", "question_count": 7, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Redox Rules", "Half-Reaction Method", "Redox Balancing" ], "new_concepts": [ "Activity Series of Metals", "Standard Cell Potential Calculation", "Voltaic Cell Design" ], "current_concepts": [ "Redox Rules", "Half-Reaction Method", "Redox Balancing", "Activity Series of Metals", "Standard Cell Potential Calculation", "Voltaic Cell Design" ] } ## Identify ionic species in unbalanced equation Using the *Redox Rules* knowledge point - $\text{MnO}_2$ is an insoluble solid oxide, so it remains as $\text{MnO}_2$. - $\text{HNO}_3$ is a strong acid, which completely dissociates into $\text{H}^+$ and $\text{NO}_3^-$. - $\text{Mn(NO}_3)_2$ is a soluble ionic salt, which dissociates into $\text{Mn}^{2+}$ and $\text{NO}_3^-$. - $\text{H}_2\text{O}$ is a molecular liquid, so it remains as $\text{H}_2\text{O}$. - Eliminating spectator ions or writing the net/ionic form yields: $$\text{MnO}_2 + \text{H}^+ + \text{NO}_3^- ightarrow \text{Mn}^{2+} + \text{NO}_3^- + \text{H}_2\text{O}$$ - This matches option (c). ## Combine half-reactions to balance redox reaction Using the *Half-Reaction Method* and *Redox Balancing* knowledge points - Reduction half-reaction: $\text{Br}_2 + 2\text{e}^- ightarrow 2\text{Br}^-$ - Oxidation half-reaction: $\text{Na} ightarrow \text{Na}^+ + \text{e}^-$ - To equalize electrons, multiply the oxidation half-reaction by 2: $$2\text{Na} ightarrow 2\text{Na}^+ + 2\text{e}^-$$ - Add the two half-reactions together: $$2\text{Na} + \text{Br}_2 ightarrow 2\text{Na}^+ + 2\text{Br}^-$$ - This matches option (c). ## Determine ease of metal oxidation - Oxidation is the loss of electrons. - Metals that lose electrons very easily are highly active metals (e.g., alkali metals). - Therefore, a highly active metal is the most easily oxidized. - This matches option (a). ## Determine ease of metal ion reduction - Reduction is the gain of electrons. The ease of reduction of a metal ion is directly related to its position in the activity series (less active metals have ions that are more easily reduced). - Among the given options (iron, mercury, aluminum, potassium), potassium and aluminum are highly active, iron is moderately active, and mercury is a noble/inactive metal. - Mercury ions ($\text{Hg}^{2+}$) have the highest reduction potential among the choices, making them the most easily reduced. - This matches option (b). ## Identify metal that plates a lead strip - For a metal to plate onto a lead ($\text{Pb}$) strip, the metal ions in solution must be reduced to their elemental form by oxidizing the lead strip. - This means lead must be more active (more easily oxidized) than the metal in solution. - Comparing activity series: $\text{Ca} > \text{Na} > \text{Mg} > \text{Al} > \text{Zn} > \text{Cr} > \text{Fe} > \text{Cd} > \text{Co} > \text{Ni} > \text{Sn} > \text{Pb} > \text{H} > \text{Cu} > \text{Hg} > \text{Ag} > \text{Au}$. - Lead ($\text{Pb}$) is more active than mercury ($\text{Hg}$), so lead will reduce mercury ions: $$\text{Pb}(s) + \text{Hg}^{2+}(aq) ightarrow \text{Pb}^{2+}(aq) + \text{Hg}(s)$$ - This matches option (c). ## Calculate cell potential of a voltaic cell - The standard cell potential ($E^\circ_{\text{cell}}$) is calculated as the reduction potential of the cathode minus the reduction potential of the anode: $$E^\circ_{\text{cell}} = E^\circ_{\text{red}} - E^\circ_{\text{oxid}}$$ - This matches option (a). ## Determine requirement for electrical energy source - To harness the flow of electrons in a redox reaction as electrical energy, the oxidation and reduction half-reactions must be physically separated (e.g., in separate half-cells connected by an external wire and a salt bridge). - This forces the electrons to travel through an external circuit to perform work. - This matches option (a). ## Question 140 a. $\text{MnO}_2 + \text{HNO}_3 ightarrow \text{Mn}^{2+} + \text{NO}_3^- + \text{H}_2\text{O}$

b. $\text{MnO}_2 + \text{H}^+ + \text{NO}_3^- ightarrow \text{MnNO}_3 + \text{H}_2\text{O}$

c. $\text{MnO}_2 + \text{H}^+ + \text{NO}_3^- ightarrow \text{Mn}^{2+} + \text{NO}_3^- + \text{H}_2\text{O}$

d. $\text{Mn}^{2+} + \text{O}^{2-} + \text{H}^+ + \text{NO}_3^- ightarrow \text{Mn}^{2+} + \text{NO}_3^- + \text{H}_2\text{O}$ ## Question 141 a. $\text{Na} + \text{Br}_2 + \text{e}^- ightarrow \text{Na}^+ + 2\text{Br}^-$

b. $\text{Na} + \text{Br}_2 ightarrow \text{Na}^+ + 2\text{Br}^-$

c. $2\text{Na} + \text{Br}_2 ightarrow 2\text{Na}^+ + 2\text{Br}^-$

d. $\text{Na} + \text{Br}_2 + 2\text{e}^- ightarrow \text{Na}^+ + 2\text{Br}^- + \text{e}^-$ ## Question 142 a. highly active metal

b. moderately active metal

c. slightly active metal

d. an inactive metal ## Question 143 a. iron

b. mercury

c. aluminum

d. potassium ## Question 144 a. potassium

b. calcium

c. mercury

d. cadmium ## Question 145 a. $E_{\text{cell}} = E_{\text{red}} - E_{\text{oxid}}$

b. $E_{\text{cell}} = E_{\text{oxid}} - E_{\text{red}}$

c. $E_{\text{cell}} = E_{\text{red}} + E_{\text{oxid}}$

d. $E_{\text{cell}} = E_{\text{red}} \times E_{\text{oxid}}$ ## Question 146 a. Two half-reactions must be physically separated.

b. One half-reaction must involve two metals.

c. Two half-reactions must involve more than one electron.

d. One half-reaction must use a metal wire electrode.

{ "subject": "Natural Science", "question_type": "Multiple Choice", "knowledge_point": [ "Natural Science", "Chemistry", "Electrochemistry" ] }

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Identify ionic species in unbalanced equation

Combine half-reactions to balance redox reaction

Determine ease of metal oxidation

Determine ease of metal ion reduction

Identify metal that plates a lead strip

Calculate cell potential of a voltaic cell

Answer:

Question 140

Question 141

Question 142

Question 143

Question 144

Question 145

Question 146