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Cell equation : $A+2 B^{+} \longrightarrow A^{2+}+2 B$ $A^{2+}+2 e^{-} \longrightarrow A ; E^{\circ}=+0.34 \mathrm{~V}$ and $\log _{10} K=15.6$ at $300 \mathrm{~K}$ for cell reactions.
Find $E^{\circ}$ for $B^{+}+e^{-} \longrightarrow B$.
$\left[\right.$ Given $: \frac{2.303 R T}{n F}=0.059$ at $\left.300 \mathrm{~K}\right]$
Options:
Find $E^{\circ}$ for $B^{+}+e^{-} \longrightarrow B$.
$\left[\right.$ Given $: \frac{2.303 R T}{n F}=0.059$ at $\left.300 \mathrm{~K}\right]$
Solution:
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Verified Answer
The correct answer is:
$0.80 \mathrm{~V}$
$\begin{aligned} & \text { } E_{\text {cell }}^{\circ}=\frac{0.059}{2} \log K \\ & E_{B^{+} / B}^{\circ}-E_{A^{2+} / A}^{\circ}=\frac{0.059}{2} \log K\end{aligned}$
$\begin{aligned} & E_{B^{+} / B}^{\circ}-0.34 \mathrm{~V}=\frac{0.059}{2} \times 15.6 \\ & E_{B^{+} / B}^{\circ}=0.46+0.34=0.80 \mathrm{~V}\end{aligned}$
$\begin{aligned} & E_{B^{+} / B}^{\circ}-0.34 \mathrm{~V}=\frac{0.059}{2} \times 15.6 \\ & E_{B^{+} / B}^{\circ}=0.46+0.34=0.80 \mathrm{~V}\end{aligned}$
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