Question

Choose the overall cell potential for this redox reaction.

Cd + Hg22+ → Cd2+ + 2Hg. Given: Cd2+ + 2e− → Cd(s) –0.40 and Hg22+ + 2e− → 2Hg(l) 0.79

A.) –0.19 V
B.) 0.19 V
C.) 0.91 V
D.) 1.19 V

Answer 1

Answer: The correct choice is D) 1.19 V.

Step-by-step explanation: Two half equations are written along with their standard reduction potential values. If we compare the standard reduction potential values then reduction of Mercury ion would be preferred as its standard reduction potential value is high. Oxidation of Cadmium will be preferred as it has less standard reduction potential value.

The over all equation also shows the same, Cadmium is oxidized and mercury is reduced.

Oxidation takes place at anode and reduction at cathode. So, anode is of cadmium and cathode is of mercury.

`E^0_c_e_l_l=E^0_(cathode)-E^0_(anode)`

`E^0_c_e_l_l=0.79V-(-0.40V)`

`E^0_c_e_l_l=0.79V+0.40V`

`E^0_c_e_l_l=1.19V`

So, the correct choice is D) 1.19 V.

Answer 2

The correct answer among all the other choices is D.) 1.19 V. This is the overall cell potential for the redox reaction Cd + Hg22+ → Cd2+ + 2Hg. Given: Cd2+ + 2e− → Cd(s) –0.40 and Hg22+ + 2e− → 2Hg(l) 0.79. Thank you for posting your question. I hope this answer helped you. Let me know if you need more help.