$$\ce{Cu^2+ + Ni -> Cu + Ni^2+}$$

The standard potential of this redox reaction is 0.57 volts (which was obtained by 0.23 + 0.34)

$$E = E_0 - \frac{RT}{nF}\ln Q$$

I used 0.1 M nickel sulphate and 0.5 M copper sulphate in my galvanic cell.

$$E_{25} = 0.57 - \frac{8.31 \cdot 298}{2 \cdot 96485}\ln(0.2)=0.59065\ldots$$

$$E_{45} = 0.57 - \frac{8.31 \cdot 318}{2 \cdot 96485}\ln(0.2)=0.59204\ldots$$

Shouldn't the two be inversely proportional? Is the effect of temperature really this minuscule?


1 Answer 1


Indeed, it should be inversely proportional. In case you divide the copper concentration by the nickel concentration, the sign of the RT/nF part should be plus and not minus. If you divide the product of the reduction by the precursor concentration, you have to use minus.

The temperature has a great effect on the current, especially in the "Cottrell-determined" area. The Nernst equation, which you use in this case, is true for a near-equilibrium environment with fast electrode kinetics. If you drive the potential too high, mass transport can become so slow (in comparison to electron transfer) that an increase in potential will not yield a higher current flow as predicted by Butler-Volmer kinetics.

  • 1
    $\begingroup$ Is there any practical significance of higher current and mass transport becoming slow? $\endgroup$ Commented Oct 10, 2016 at 22:44
  • $\begingroup$ It depends what you are looking for. You can either work at low potentials, have slow rate constants but mass transport will always 'bring enough reactants to the electrode'. At high potentials, the rate constants are fast enough, but mass transport is too slow and what is near the electrode will be quickly deposited. In electrodeposition/electroplating, working at low potentials (in the Nernst-controlled region) usually leads to more homogeneous electrodeposited films. $\endgroup$ Commented Oct 10, 2016 at 22:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.