Consider the following cell: Pt | Fe 3+ (2.00 x 10 -2 M), Fe 2+ (6.00 x 10 -5 M)

Question

Consider the following cell:

Pt | Fe3+ (2.00 x 10-2 M), Fe2+ (6.00 x 10-5 M) || Sn2+ (3.5 x 10-2 M), Sn4+(1.5 x 10-4 M) | Pt

A) Is the cell (as written) galvanic or electrolytic? Support with calculations.

B) What is the anode? Cathode?

C) Write a balanced equation for the cathodic reaction.

D) Write a balanced equation for the anodic reaction.

E) Write a balanced chemical equation for the cell (as written).

F) What is Ecell?

Fe3++ e- >> Fe2+   +0.771 v

Fe2+ + 2e- >> Fe    -0.440 v

Sn2+ + 2e- >> Sn    -0.136 v

Sn4+ + 2e- >> Sn2+ +0.154 v

Please show work for all parts!

Explanation / Answer

I will answer all parts together. When the cell is written with this format: "X/Y//Z/W", ALWAYS at the left side is the anode and at the rigth side is the cathode.

(Fe+2 ------> Fe+3 + e-)x2 anode (oxidation) E0=0.771V

Sn+4 + 2e- -----> Sn+2 cathode (reduction) E0= 0.154V

----------------------------------------------------------------------

2Fe+2 + Sn+4 -----> 2Fe+3 + Sn+2 E0cell= E0cathode- E0anode= 0.154-0.771= -0.617V

We can´t tell if the cell is galvanic or electrolytic looking at E0cell, this is because the system is not at standard conditions. Because of that we have to calculate Ecell in order to know how the concentration of the ions affect the potential. If the potential is negative, then the cell is electrolytic, if it is positive, is galvanic.

Ecell= E0cell - 0.059/n log Q

n= number of electrons exchanged, in this case 2

Ecell= -0.617V -0.059/2 log [Sn+2][Fe+3]2/[Fe+2]2[Sn+4]

Ecell= -0.617 -0.059/2 log(0.035)(0.02)2/(6x10-5)2(1.5x10-4)= -0.836V ----> The cell is electrolytic

I hope it helps, let me know if you have any doubt :)

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