Calculate the E_{cell} for Zn_{(s)} | Zn^{2+} | vedclass

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Question

(B) The cell reaction is: $3 Zn_{(s)} + 2 Cr^{3+}_{(0.1 \ M)} \longrightarrow 3 Zn^{2+}_{(0.1 \ M)} + 2 Cr_{(s)}$Using the Nernst equation: $E_{cell}

Vedclass · Accepted Answer

$0.03$

Vedclass · Answer

(B) The cell reaction is: $3 Zn_{(s)} + 2 Cr^{3+}_{(0.1 \ M)} \longrightarrow 3 Zn^{2+}_{(0.1 \ M)} + 2 Cr_{(s)}$Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.0592 \ V}{n} \log_{10} \frac{[Zn^{2+}]^3}{[Cr^{3+}]^2}$Here,$n = 6$ (total electrons transferred).$E_{cell} = 0.02 \ V - \frac{0.0592 \ V}{6} \log_{10} \frac{(0.1)^3}{(0.1)^2}$$E_{cell} = 0.02 \ V - \frac{0.0592 \ V}{6} \log_{10} (0.1)$Since $\log_{10} (0.1) = -1$,we have:$E_{cell} = 0.02 \ V - \frac{0.0592 \ V}{6} 	imes (-1)$$E_{cell} = 0.02 \ V + 0.00987 \ V \approx 0.03 \ V$.

Calculate the $E_{cell}$ for $Zn_{(s)} | Zn^{2+}_{(0.1 \ M)} || Cr^{3+}_{(0.1 \ M)} | Cr_{(s)}$ at $25^{\circ} C$ if $E^{\circ}_{cell}$ is $0.02 \ V$. (in $V$)

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