Q54 / 180 · AIIMS 2018 (26 May) · Chemistry

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 $\left.300 \mathrm{~K}\right]$

  • Correct answer
    A
    $0.80 \mathrm{~V}$
  • B
    $1.26 \mathrm{~V}$
  • C
    $-0.54 \mathrm{~V}$
  • D
    $+0.94 \mathrm{~V}$

Solution

$\begin{aligned} & \text { } E_{\text {cell }}^{\circ}=\frac{0.059}{2} \log K \\ & E_{B^{+} / B}^{\circ}-E_{A^{2+} / A}^{\circ}=\frac{0.059}{2} \log K\end{aligned}$ $\begin{aligned} & E_{B^{+} / B}^{\circ}-0.34 \mathrm{~V}=\frac{0.059}{2} \times 15.6 \\ & E_{B^{+} / B}^{\circ}=0.46+0.34=0.80 \mathrm{~V}\end{aligned}$

Hence, the correct answer is option: 'A'.

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