Calculate the emf of the following cell at 25°C: $$\mathrm{Ag(s)\,|\,Ag^+(10^{-3}\ M)\,||\,Cu^{2+}(10^{-1}\ M)\,|\,Cu(s)}$$ Given: $E^{\circ}_{cell} = +0.46$ V and log 10$^n$ = n.

CBSE Class 12 · 2013 · Redox Reactions and ElectrochemistryLong Answer

Calculate the emf of the following cell at 25°C:

Ag (s) ∣ A g^{+} (1 0^{- 3} M) ∣∣ C u^{2 +} (1 0^{- 1} M) ∣ Cu (s)

Given:

E_{ce l l}^{\circ} = + 0.46

V and log 10

^{n}

= n.

Answer: E_cell ≈ 0.61 V (*)

Solution

As written, Ag is the anode: 2Ag + Cu

^{2 +} \to

2Ag

^{+}

+ Cu,

n = 2

, with the given

E_{ce l l}^{\circ} = 0.46

E_{ce l l} = 0.46 - \frac{0.059}{2} lo g \frac{[ A g ^{+} ] ^{2}}{[ C u ^{2 +} ]} = 0.46 - 0.0295 lo g \frac{1 0 ^{- 6}}{1 0 ^{- 1}}

= 0.46 - 0.0295 \times (- 5) = 0.46 + 0.1475

E_{ce l l} = 0.6075 \approx 0.61 V

(*) The cell as printed pairs with the given E° only in this sense (the 2013 paper's intended computation); physically the spontaneous cell runs the other way with Cu as anode.

CBSE Class 12 Board · 2013 · Subjective

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CBSE Class 12 · 2013 · Redox Reactions and ElectrochemistryLong Answer

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