Ncert Exercises & Solutions

Mg_(s) | Mg²⁺_(0.001M) || Cu²⁺_{(0.0001 M)} | Cu_(s)

$E_{{Mg}^{2 +} / Mg}^{o} = - 2.36 V; E_{{Cu}^{2 +} / Cu}^{o} = 0.34 V$

The value of $E_{cell}$ for the above reaction is -

1. 3.46 V

2. 3.15 V

3. 2.67 V

4. 1.24 V

Hint:

E_{cell} = E_{cell}^{⊝} - \frac{0.0591}{n} \log \frac{[{Mg}^{2 +}]}{[{Cu}^{2 +}]}

Step 1:

Calculate the value of

E_{cell}^{o}

as follows:

E_{cell}^{⊝} = E_{Mg / {Mg}^{2 +}}^{o} + E_{{Cu}^{2 +} / Cu}^{o}

E_{cell}^{⊝} = 2.36 + 0.34 V

E_{cell}^{⊝} = 2.70 V

Step 2:

Calculate the value of

E_{cell}

as follows:

$E_{cell} = E_{cell}^{⊝} - \frac{0.0591}{n} \log \frac{[{Mg}^{2 +}]}{[{Cu}^{2 +}]}$

$= {0.34 - (- 2.36)} - \frac{0.0591}{2} \log \frac{. 001}{. 0001}$

$= 2.7 - \frac{0.0591}{2} \log 10$

=2.7 - 0.02955

=2.67 V (approximately)