A student measures the cell EMF, EcellE_{\text{cell}}Ecell, for the electrochemical cell represented by:
Ni(s)∣Ni2+(aq)∣∣Pb2+(aq)∣Pb(s) \text{Ni(s)}|\text{Ni}^{2+}(\text{aq})||\text{Pb}^{2+}(\text{aq})|\text{Pb(s)} Ni(s)∣Ni2+(aq)∣∣Pb2+(aq)∣Pb(s)at temperature TTT.
The student plots a graph of EcellE_{\text{cell}}Ecell against ln([Ni2+][Pb2+])\ln \left( \frac{[\text{Ni}^{2+}]}{[\text{Pb}^{2+}]} \right)ln([Pb2+][Ni2+]) and draws a line of best fit. Two points on this line of best fit are:
Calculate the gradient of this line of best fit. Show your working.
Use your gradient to calculate the temperature, TTT, in Kelvin, at which these measurements were taken, using the relation:
Ecell=(−4.3×10−5×T)ln([Ni2+][Pb2+])+Ecellθ E_{\text{cell}} = (-4.3 \times 10^{-5} \times T) \ln \left( \frac{[\text{Ni}^{2+}]}{[\text{Pb}^{2+}]} \right) + E^\theta_{\text{cell}} Ecell=(−4.3×10−5×T)ln([Pb2+][Ni2+])+EcellθIn one specific experiment, where [Pb2+]=1.0 mol dm−3[\text{Pb}^{2+}] = 1.0\text{ mol dm}^{-3}[Pb2+]=1.0 mol dm−3 and [Ni2+]=0.10 mol dm−3[\text{Ni}^{2+}] = 0.10\text{ mol dm}^{-3}[Ni2+]=0.10 mol dm−3, the measured cell EMF is 0.14 V0.14\text{ V}0.14 V. Under these conditions, the electrode potential of the Pb2+/Pb\text{Pb}^{2+}/\text{Pb}Pb2+/Pb electrode is −0.13 V-0.13\text{ V}−0.13 V. Calculate the electrode potential for the Ni2+/Ni\text{Ni}^{2+}/\text{Ni}Ni2+/Ni electrode.
Give one reason why this calculated value for the electrode potential is different from the standard electrode potential of the Ni2+/Ni\text{Ni}^{2+}/\text{Ni}Ni2+/Ni electrode (−0.25 V-0.25\text{ V}−0.25 V).
