There is an electrolytic cell in which Mn2+ is reduced to Mn and Sn is oxidized to Sn2+. A. Write an equation for the half-reaction occurring at each electrode. Express your answers as chemical equations separated by a comma. Identify all of the phases in your answer.
B. What minimum voltage is necessary to drive the reaction? Vmin =

-1.05 V

Explanation:

At the anode;

Sn(s) > Sn^2+(aq) + 2e

At the cathode;

Mn^2+(aq) + 2e > Mn(s)

Given that it is an electrolytic cell,

E°cell= E°cathode - E°anode

But;

E°cathode= -1.19 V

E°anode= -0.14 V

E°cell= -1.19-(-0.14)

E°cell= -1.05 V

Therefore

Vmin = -1.05 V

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.

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