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CBSE Class 12 Chemistry 2023 Delhi Set 2 Solved Paper

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Question : 1 of 12
Marks: +1, -0
SECTION - A
Consider the following standard electrode potential values:
  Fe(aq)3++e−→Fe(aq)2+E∘\; \mathrm{Fe}^{3+}_{\text{(aq)}} + e^{-} \rightarrow \mathrm{Fe}^{2+}_{\text{(aq)}} E^{\circ} =+0.77V= +0.77 \mathrm{V}
  MnO4  (aq)  −+8H++5e−→Mn2+  (aq)  \; {\mathrm{MnO}_4}^{-}_{\;\text{(aq)}\;} + 8 \mathrm{H}^{+} + 5 e^{-} \rightarrow {\mathrm{Mn}^{2+}}_{\;\text{(aq)}\;} +4H2O(l)E∘=++4 \mathrm{H}_2 \mathrm{O}_{\text{(l)}} E^{\circ}=+
  1.51V\;1.51 \mathrm{V}
What is the cell potential for the redox reaction
Solution:     👈: Video Solution
E∘=EC−EaE^{\circ} = E_{C} - E_{a}
=1.51−0.77=+0.74V=1.51-0.77=+0.74 \mathrm{V}
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