Question

decide whether each chemical reaction in the table below is an oxidation - reduction (
edox\) reaction. if the reaction is a redox reac of the reducing agent and the formula of the oxidizing agent.
$3cr(no_{3})_{2}(aq)+2al(s)\to3cr(s)+2al(no_{3})_{3}(aq)$
$hso_{4}^{-}(aq)+oh^{-}(aq)\to so_{4}^{2 - }(aq)+h_{2}o(l)$
$2cl_{2}(g)+sn(s)\to sncl_{4}(s)$

Explanation:

Step1: Analyze the first reaction \(3Cr(NO_3)_2(aq)+2Al(s)\to3Cr(s)+2Al(NO_3)_3(aq)\)

• For \(Al\): \(Al\) goes from \(0\) (in \(Al(s)\)) to \(+ 3\) (in \(Al(NO_3)_3\)). It is oxidized, so \(Al\) is the reducing agent.
• For \(Cr\): \(Cr\) goes from \(+2\) (in \(Cr(NO_3)_2\)) to \(0\) (in \(Cr(s)\)). It is reduced, so \(Cr(NO_3)_2\) is the oxidizing agent. This is a redox reaction.

Step2: Analyze the second reaction \(HSO_4^-(aq)+OH^-(aq)\to SO_4^{2 - }(aq)+H_2O(l)\)

• Calculate oxidation numbers:
• In \(HSO_4^-\): \(H = + 1\), \(O=-2\), let \(S=x\), then \(+1+x+4\times(-2)=-1\), \(x = + 6\).
• In \(SO_4^{2-}\): \(O=-2\), let \(S = y\), then \(y + 4\times(-2)=-2\), \(y=+6\).
• In \(OH^-\): \(O=-2\), \(H = + 1\).
• In \(H_2O\): \(O=-2\), \(H=+1\).
• No change in oxidation numbers. So it is not a redox reaction.

Step3: Analyze the third reaction \(2Cl_2(g)+Sn(s)\to SnCl_4(s)\)

• For \(Sn\): \(Sn\) goes from \(0\) (in \(Sn(s)\)) to \(+4\) (in \(SnCl_4\)). It is oxidized, so \(Sn\) is the reducing agent.
• For \(Cl\): \(Cl\) goes from \(0\) (in \(Cl_2\)) to \(-1\) (in \(SnCl_4\)). It is reduced, so \(Cl_2\) is the oxidizing agent. This is a redox reaction.

Answer:

• For \(3Cr(NO_3)_2(aq)+2Al(s)\to3Cr(s)+2Al(NO_3)_3(aq)\): redox reaction? yes; reducing agent: \(Al\); oxidizing agent: \(Cr(NO_3)_2\)
• For \(HSO_4^-(aq)+OH^-(aq)\to SO_4^{2 - }(aq)+H_2O(l)\): redox reaction? no; reducing agent: (not applicable); oxidizing agent: (not applicable)
• For \(2Cl_2(g)+Sn(s)\to SnCl_4(s)\): redox reaction? yes; reducing agent: \(Sn\); oxidizing agent: \(Cl_2\)