Question

Sn4+ (1.50 M) +Zn (s) ----> Sn2+ (0.50M) + Zn2+
i. Calculate the potential of the following cell reaction at 298 K.
ii.The standard potential E of the cell is 0.89 V. Whether the potential of cell will increase or decrease if concentration of Sn4+ is increased in the cell

Answer 1

Hi, you need to note that the values are note unity and thus we use the Nerst equation to calculate the potential of the reaction Sn4+ (1.50 M) +Zn (s) ----> Sn2+ (0.50M) + Zn2+

Ecell = E° - O.O592/n log [Sn2+] [Zn2+]/[Sn4+]

Ecell= 0.89 - 0.0592/2 log (0.5 x 1)/1.5

Ecell = 0.9 V

Answer 2

Answer:

Explanation:

Sn4+ + Zn → Sn2+ + Zn2+

At anode : Zn → Zn2++ 2e−

At cathode : Sn4+ + 2e− → Sn2+

Here, n=2

(i) E=E°−0.059nlog [Sn2+][Sn4+]    

  =0.89 +0.0592log0.501.50

        =0.89+0.0592×(−0.4771)        

= 0.89−0.0140  

     =0.87 V

(ii) If the concentration of Sn4+ will increase then the ratio of [Sn2+][Sn4+] will decrease and so the log value of this ratio will increase and overall, the potential of the cell will decrease.