Question

question 18.c of 22
consider a galvanic electrochemical cell constructed using cr/cr³⁺ and zn/zn²⁺ at 25 °c. the following half - reactions are provided for each metal:
cr³⁺(aq) + 3 e⁻ → cr(s) e°red = -0.744 v
zn²⁺(aq) + 2 e⁻ → zn(s) e°red = -0.763 v
write the balanced equation for the overall reaction in acidic solution.

Explanation:

Step1: Determine the oxidation and reduction half - reactions

The more positive \(E^{\circ}_{red}\) value indicates the reduction half - reaction. Since \(- 0.744\ V> - 0.763\ V\), \(Cr^{3+}(aq)+3e^{-}\to Cr(s)\) is the reduction half - reaction. Then \(Zn(s)\to Zn^{2 +}(aq)+2e^{-}\) is the oxidation half - reaction (reverse of \(Zn^{2+}(aq)+2e^{-}\to Zn(s)\)).

Step2: Balance the electrons

For the reduction half - reaction \(Cr^{3+}(aq)+3e^{-}\to Cr(s)\), and the oxidation half - reaction \(Zn(s)\to Zn^{2 +}(aq)+2e^{-}\). The least common multiple of \(2\) and \(3\) (the number of electrons) is \(6\).
Multiply the reduction half - reaction by \(2\): \(2Cr^{3+}(aq)+6e^{-}\to 2Cr(s)\)
Multiply the oxidation half - reaction by \(3\): \(3Zn(s)\to 3Zn^{2 +}(aq)+6e^{-}\)

Step3: Combine the two half - reactions

Add the two balanced half - reactions together:
\(2Cr^{3+}(aq)+3Zn(s)\to 2Cr(s)+3Zn^{2 +}(aq)\)

Answer:

\(2Cr^{3+}(aq)+3Zn(s)\to 2Cr(s)+3Zn^{2 +}(aq)\)