AP EAMCET · Chemistry · Electrochemistry
The electrode potential for
\[
\begin{aligned}
& M^{2+}(a q)+e^{-} \longrightarrow M^{+}(a q) \\
& M^{+}(a q)+e^{-} \longrightarrow M(s)
\end{aligned}
\]
are \(+0.15 \mathrm{~V}\) and \(+0.50 \mathrm{~V}\) respectively. The value of \(E^{\circ}{ }_{M^{2+} / M}\) will be
- A 0.150 V
- B 0.300 V
- C 0.325 V
- D 0.650 V
Answer & Solution
Correct Answer
(C) 0.325 V
Step-by-step Solution
Detailed explanation
Given, \[ \begin{aligned} & M^{2+}(a q)+e^{-} \longrightarrow M^{+}(a q), E_1^{\circ}=+0.15 \mathrm{~V} \\ & M^{+}(a q)+e^{-} \longrightarrow M(s), E_2^{\circ}=0.5 \mathrm{~V} \end{aligned} \] \(\because\) Free energy, i.e. \(\left[\Delta G^{\circ}\right]=-n F E^{\circ}\) where,…
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