Answer:
[tex]1.66~V[/tex]
Explanation:
We have to start with the half-reactions for both ions:
[tex]Zn^+^2~+2e^-~->Zn[/tex] V= -0.76
[tex]Ag^+~e^-~->~Ag[/tex] V= +0.80
If we want a spontaneous reaction (galvanic cell) we have to flip the first reaction, so:
[tex]Zn~->~Zn^+^2~+2e^-~[/tex] V= +0.76
[tex]Ag^+~+~e^-~->~Ag[/tex] V= +0.80
If we want to calculate ºE we have to add the two values, so:
ºE=0.76+0.80 = 1.56 V
Now, we have different concentrations. So, if we want to calculate E we have to use the nerts equation:
[tex]E=ºE~+~\frac{0.059}{n}LogQ[/tex]
On this case, Q is equal to:
[tex]Q=\frac{[Zn^+^2]}{[Ag^+]^2}[/tex]
Because the total reaction is:
[tex]Zn~+~2Ag^+~->~Zn^+^2~+~2Ag[/tex]
So, the value of "Q" is:
[tex]Q=\frac{[0.052 M]}{[0.0042]^2}=2947.84[/tex]
Now, we can plug all the values in the equation (n=2, because the amount of electrons transferred is 2). So:
[tex]E=1.56~V~+~\frac{0.059}{2}Log(2947.84)=1.66~V[/tex]
I hope it helps!
