Question

Write cell reaction and calculate the potential (EMF) of the cell Zn(s) | Zn²⁺(0.08 M) || Cr³⁺ (0.1 M) Cr(s) at 298 K. 
\[\ce{E^0_{Zn} = - 0.76 V and E^0_Cr = - 0.74 V}\]

Answer 1

Given: [Zn²⁺] = 0.08 M; [Cr³⁺] = 0.1 M

\[\ce{E^0_{{{Zn}^{2+}}/Zn} = - 0.76 V; E^0_{{{Cr}^{3+}}/Cr} = - 0.74 V}\]

\[\ce{E^0_{cell} = ?}\]

Cell reaction:

(LHS) 3Zn(s) → 3Zn²⁺(aq) + 6e^-

(RHS) 2Cr³⁺(aq) + 6e^- → 2Cr(s)
___________________________

Net cell reaction 3Zn(s) + 2Cr³⁺(aq) → 3Zn²⁺(aq) + 2Cr(s)
                                          0.1M             0.08 M       (n = 6)

\[\ce{E^0_{cell} = E^0_{{{Cr}^{3+}}/Cr} - E^0_{{{Zn}^{2+}}/Zn}}\]

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

`E_(cell) = E_(cell)^0 - 0.0592/n  log_10  [Zn^(2+)]^3/[Cr^(3+)]^2`

`= 0.02 - 0.0592/n  log_10  (0.08)^3/(0.1)^2`

`0.02 - 9.867 xx 10^-3 log_10  (5.12 xx 10^-4)/(1 xx 10^-2)`

`= 0.02 - 9.867 xx 10^-3 xx bar2. 7093`

`= 0.02 - 9.867 xx 10^-3 xx (- 1.2907)`

`= 0.02 + 0.01273`

`= 0.03273 V`