WBJEE · Chemistry · Electrochemistry
\(\begin{array}{lllll}\text {The two } & \text { half-cell } & \text { reactions } & \text { of } & \text { an }\end{array}\) electrochemical cell is given as \(\mathrm{Ag}^{+}+e^{-} \longrightarrow \mathrm{Ag} ; \quad E_{\mathrm{Ax}^{+} / \mathrm{Ag}}^{\circ}=-0.3995 \mathrm{V}\) \(\mathrm{Fe}^{3+} \longrightarrow \mathrm{Fe}^{3 *}+e^{-} ; E_{\mathrm{Fe}^{3+} \mathrm{Fv}^{2+}}=-0.7120 \mathrm{V}\). The value of cell EMF will be
- A \(-0.3125 \mathrm{V}\)
- B \(0.3125 \mathrm{V}\)
- C \(1.114 \mathrm{V}\)
- D \(-1.114 \mathrm{V}\)
Answer & Solution
Correct Answer
(B) \(0.3125 \mathrm{V}\)
Step-by-step Solution
Detailed explanation
Species with more negative \(E^{\circ}\) (standard reduction potential) generally acts as reducing agent while with less negative \(E^{\circ}\) acts as oxidising agent. Thus, the overall reaction is…
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