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19. Given the following cell diagram in acidic medium and at 25 ºc:rnmno2(s) mn2+(aq) ce4+(aq), Ce3+(aq)rnd) Calculate the ph at which the electrochemical cell is able to generate a differencernof potential of +0.65 v.
Accepted Answer
d) At 25 °C, the Nernst equation is given by:
E°cell = E°cell - (0.0592 V / n) * log Q
Where:
- E°cell is the standard cell potential
- E°cell is the cell potential under nonstandard conditions
- n is the number of electrons transferred in the balanced chemical equation
- Q is the reaction quotient
In this case, the standard cell potential is +1.51 V, n = 2, and Q = [Ce4+][Ce3+]/[Mn2+].
We can use the Nernst equation to calculate the concentration of Mn2+ at which the cell potential is +0.65 V:
0.65 V = 1.51 V - (0.0592 V / 2) * log ([Ce4+][Ce3+]/[Mn2+])
Solving for [Mn2+], we get:
[Mn2+] = [Ce4+][Ce3+]/10^14.5
To calculate the pH at which this concentration of Mn2+ is present, we can use the following equation:
[H+] = [Mn2+] * 10^-10.5
Substituting the expression for [Mn2+] from above, we get:
[H+] = [Ce4+][Ce3+] * 10^-4.5
Taking the negative logarithm of both sides, we get:
-log[H+] = -log([Ce4+][Ce3+]) + 4.5
-log[H+] = pH = 4.5 - log([Ce4+][Ce3+])
Therefore, the pH at which the electrochemical cell is able to generate a difference of potential of +0.65 V is:
pH = 4.5 - log([Ce4+][Ce3+])
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