Determine the potential for the cell:
$P t | F e^{2}, F e^{3 +} | | C r_{2} O_{7}^{2 +}, C r^{3 +}, H^{+} | P t$
in which $[F e^{2 +}]$ and $[F e^{3 +}]$ are $0.5 M$ and $0.75 M$ respectively and $[C r_{2} O_{7}^{2 -}], [C r^{3 +}]$ and $[H^{+}]$ are $2 M, 4 M$ and $1 M$ respectively, Given:
$F e^{3 +} + e^{-} \to F e^{2 +}; E^{\circ} = 0.770 v o l t$
$14 H^{+} + 6 e^{-} + C r_{2} O_{7}^{2 -} \to 2 C r^{3 +}, 7 H_{2} O; E^{\circ} = 1.35 v o l t$ .

0.26 v o l t

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0.36 v o l t

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0.46 v o l t

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0.58 v o l t

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Solution

The correct option is C $0.58 v o l t$
potential cell = cathode RP - anode RP
cathode RP = 1.35
anode RP = 0.770
so equal 1.35 - 0.77 = 0.58
D is correct answer

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