Question

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 given that the standard cell potential is 0.019 v and the overall equation is 2 cr³⁺(aq) + 3 zn(s) → 2 cr(s) + 3 zn²⁺(aq), what is the cell potential for this cell at 25.0 °c when zn²⁺ = 0.0143 m and cr³⁺ = 0.1556 m?

Explanation:

Step1: Identify the Nernst equation

The Nernst equation is \(E = E^{\circ}-\frac{RT}{nF}\ln Q\). At \(25^{\circ}C\) (\(T = 298\ K\)), \(R = 8.314\ J/(mol\cdot K)\), \(F=96485\ C/mol\), and \(\frac{RT}{F}=0.0257\ V\). For the reaction \(2Cr^{3 +}(aq)+3Zn(s)\to2Cr(s)+3Zn^{2 +}(aq)\), \(n = 6\) (from the number of electrons transferred in the balanced half - reactions: \(2\times3e^{-}=6e^{-}\) or \(3\times2e^{-}=6e^{-}\)). The reaction quotient \(Q=\frac{[Zn^{2 +}]^{3}}{[Cr^{3 +}]^{2}}\).

Step2: Substitute values into the Nernst equation

Substitute \(E^{\circ}=0.019\ V\), \(n = 6\), \([Zn^{2 +}]=0.0143\ M\), and \([Cr^{3 +}]=0.1556\ M\) into the Nernst equation.
First, calculate \(Q\):
\(Q=\frac{(0.0143)^{3}}{(0.1556)^{2}}=\frac{2.924\times10^{-6}}{0.0242}=1.208\times10^{-4}\)
Then, \(\ln Q=\ln(1.208\times 10^{-4})=-8.91\)
Now, substitute into \(E = E^{\circ}-\frac{RT}{nF}\ln Q\):
\(E=0.019-\frac{0.0257}{6}\times(- 8.91)\)
\(E = 0.019+0.038\)

Answer:

\(0.057\ V\)