Verified 30 May 2026.
For the cell $\text{Zn}|\text{Zn}^{2+}(0.01\text{M})||\text{Cu}^{2+}(1\text{M})|\text{Cu}$, if $E^{\circ}_{\text{cell}} = 1.1\text{V}$, the EMF at $298\text{K}$ is:
$(R = 8.314 \text{ J mol}^{-1}\text{K}^{-1},\; F = 96500 \text{ C mol}^{-1})$
$1.1592 \text{ V}$
$1.0408 \text{ V}$
$1.1 \text{ V}$
$1.2184 \text{ V}$
Using the Nernst equation:
$$E = E^{\circ} - \frac{RT}{nF}\ln Q = E^{\circ} - \frac{0.0592}{n}\log Q$$
For $\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$, $n = 2$
$$Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{0.01}{1} = 0.01$$
$$E = 1.1 - \frac{0.0592}{2}\log(0.01) = 1.1 - 0.0296 \times (-2)$$
$$= 1.1 + 0.0592 = \boxed{1.1592 \text{ V}}$$
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
For an endothermic reaction at equilibrium, increasing temperature will:
For the reaction N₂O₄(g) ⇌ 2NO₂(g), if Kp = 4atm at 300K, the degree of dissociation at a total pressure of 1 atm is:
For the reaction N₂ + 3H₂ ⇌ 2NH₃, if Kp = 0.04 atm⁻² at a certain temperature, the value of Kc in terms of R and T is: