The measured voltage for the reaction with the indicated concentrations is $1.5 V$ . Calculate $E_{cell}^{\circ} in V$ .
$C r (s) + 3 A g^{+} (a q, 0.1 M) \to 3 A g (s) + C r^{3 +} (a q, 0.1 M)$

1.54

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1.65

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1.35

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1.46

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Solution

The correct option is A $1.54$
$C r (s) + 3 A g^{+} (a q, 0.1 M) \to 3 A g (s) + C r^{3 +} (a q, 0.1 M)$
Here number of electrons involved $= 3$
Here $E_{cell} = E_{cell}^{\circ} - \frac{0.0591}{3} log \frac{[C r^{3 +}]}{[A g^{+}]^{3}} \Rightarrow E_{cell}^{\circ} = 1.5 + \frac{0.0591}{3} log \frac{0.1}{10^{- 3}} \Rightarrow E_{cell}^{\circ} = 1.5 + \frac{0.0591}{3} log 100 \Rightarrow E_{cell}^{\circ} = 1.5 + \frac{0.0591}{3} \times 2 \Rightarrow E_{cell}^{\circ} = 1.54 V$

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Similar questions

For the galvanic cell :

A l (s) | A l^{3 +} (a q) (0.1 M) | A g^{+} (a q) (0.1 M) | A g (s),

which of the following is/are correct?

^{\circ}

^{\circ}

Δ

^{\circ}

C r (s) + 3 A g^{+} (a q, 0.1 M) \to A g (s) + C r^{3 +} (a q, 0.1 M)

N i (s) | N i^{2 +} (a q . 0.1 M) | | A g^{+} (a q . 0.1 M | A g (s)

E_{A g^{+} / A g}^{\circ} = 0.80 v o l t

E_{N i^{2 +} / N i}^{\circ} = - 0.25 v o l t

Δ G^{o}

25^{o}

Z n (s) | Z n_{a q}^{2 +} (0.1 M) | C d_{a q}^{2 +} (0.01 M) | C d_{(s)}

E^{o} Z n^{2 +} / Z n = - 0.763

E^{o} C d^{2 +} / C d = - 0.403 V .

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