Single Choice

Calculate the standard cell potential in (V) of the cell in which following reaction takes place: $$Fe^{2+}(aq) + Ag^+(aq) \rightarrow Fe^{3+}(aq) + Ag(s)$$ Given that $$E^o_{Ag^+/Ag} = xV$$ $$E^o_{Fe^{2+}/Fe} = yV$$ $$E^o_{Fe^{3+}/Fe} = zV$$

A$$x + 2y - 3z$$
Correct Answer
B$$x - z$$
C$$x - y$$
D$$x + y - z$$

Solution

Solution:- (A) $$x + 2y - 3z$$

$$Fe^{2+}(aq) + Ag^{+}(aq) \rightarrow Fe^{3+} (aq) + Ag(s)$$

Cell reaction,

Anode: $$Fe^{2+}(aq) \rightarrow Fe^{3+} (aq) + e^{\ominus}; E^o_{Fe^{2+}/Fe^{3+}} = mV$$

Cathode: $$Ag^+(aq) + e^{\ominus} \rightarrow ; Ag(s); E^o_{Ag^+/Ag} = xV$$

$$\Rightarrow$$ cell standard potential $$= (m + x)V$$

Now, to find 'm',

$$Fe^{2+}+2e^{\ominus} \rightarrow Fe; E^o_1 = yV \Rightarrow \Delta G_1^o = (2Fy)$$

$$Fe^{3+}+3e^{\ominus} \rightarrow Fe; E^o_2 = zV \Rightarrow \Delta G_2^o = (3Fy)$$

$$Fe^{2+}(aq) \rightarrow Fe^{3+}(aq) + e^{\ominus}; E^o_3 = mV \Rightarrow \Delta G_3^o = -(1Fm)$$

$$\Delta G_3^o = \Delta G^0_1 - \Delta G^o_2 = (-2Fy + 3Fz) = -Fm$$

$$\Rightarrow m = (2y - 3z)$$

$$\Rightarrow E^o_{cell} = (x + 2y - 3z)V$$

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