A) increases cell emf
B) increases the current flow
C) decrease the cell emf
D) alter the pH of the solution
Correct Answer: C
Solution :
\[2F{{e}^{3+}}+zn\to Z{{n}^{2+}}+2F{{e}^{2+}}\] \[2F{{e}^{3+}}+2{{e}^{-}}\to 2F{{e}^{2+}}\] According Nemsts equation, \[{{E}_{cell}}=E_{cell}^{o}-\frac{RT}{nF}\ln \frac{{{[F{{e}^{2+}}]}^{2}}}{{{[F{{e}^{3+}}]}^{2}}}\] \[{{\text{E}}_{\text{cell}}}\]is depend on concentration of \[\text{F}{{\text{e}}^{2+}}.\] Therefore, \[[F{{e}^{2+}}]\]decrease \[{{E}_{cell}}.\]
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