For the following electrochemical cell at 298K, Pt(s)|H2(g,1bar)|H+(aq,1M)||M4+(aq),M2+(aq)|Pt(s), Ecell=0.092V when [M2+(aq)]/[M4+(aq)]=10x. Given: E0M4+/M2+=0.151V; 2.303RT/F=0.059V. The value of x is :

Given:
Ecell = 0.092 V
E°M4+/M2+ = 0.151 V
2.303RT/F = 0.059 V
[M2+(aq)]/[M4+(aq)] = 10x
The Nernst equation for the given cell is:
Ecell = E°cell - (2.303RT/nF)logQ
where Q is the reaction quotient.
For the given cell reaction:
M4+(aq) + 2e- ⇌ M2+(aq)
The number of electrons transferred (n) is 2.
The reaction quotient Q is given by:
Q = [M2+(aq)]/[M4+(aq)] = 10x
Substituting the given values in the Nernst equation:
0.092 = 0.151 - (0.059/2)log(10x)
0.092 - 0.151 = - (0.059/2)log(10x)
-0.059 = - (0.059/2)log(10x)
log(10x) = 2
10x = 10^2
10x = 100
x = 10
Therefore, the value of x is 10.