vi. Calculate emf of the following cell at 250C. Zn (s) |Zn2⊕(0.08M)||Cr3⊕(0.1M)|Cr E0 Zn = - 0.76 V, E0 Cr = - 0.74 V (0.0327 V)

chapter 5 electrochemistry class 12 chemistry textbook solution

vi. Calculate emf of the following cell at 250C. Zn (s) |Zn2⊕(0.08M)||Cr3⊕(0.1M)|Cr E0
Zn = – 0.76 V, E0 Cr = – 0.74 V (0.0327 V)

Answer:-

Given:

E_{Zn}^{\circ} = - 0.76 V, E_{Cr}^{\circ} = - 0.74 V

To find: Emf of the cell (E_{cell})

Formulae:

E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}

E_{cell} = E_{cell}^{\circ} - \frac{0.0592 V}{n} \log_{10} [\frac{Product}{Reactant}]

Calculation:

3 {Zn}_{(s)} \to 2 {Cr}_{(0.1 M)}^{3 +} - 3 {Zn}_{(0.08 M)}^{2 +} + {Cr}_{(s)} (overall reaction)

Using formula (1):

E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}

E_{cell} = E_{Cr}^{\circ} - E_{Zn}^{\circ}

= - 0.74 V - (- 0.76 V) = 0.02 V

Using formula (2):

The cell potential is given by

E_{cell} = E_{cell}^{\circ} - \frac{0.0592 V}{6} \log_{10} (\frac{{0.08}^{3}}{{0.1}^{2}})

= 0.02 V + 0.0127 = 0.0327 V

The emf of the cell is

0.0327 V

chapter 5 elctrochemistry textbook solution page 118