Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Zn2 ] = 0.821 M and [Ni2 ] = 0.0200 M. Standard reduction potentials can be found here.
Zn(s) + Ni+2(aq)=Zn+2(aq) + Ni(s)
we have use the equation
E cell = E0 cell - 0.059/n logQ
first find out the E0 cell using standard reduction potential values
Ni^2+ + 2e- --> Ni(s) E0: -0.28V
Zn^2+ + 2e- --> Zn(s) E0: -0.763V
Zn+ Ni^2+ --> Zn^2+ Ni .. E0: -0.28 -(-0.76) = 0.48V
this is the E0 of the cell
Q = [Zn2+] / [Ni2+] = 0.821 M / 0.02 M = 41.05
log Q = log(41.05) = 1.61
n= no of electrons involved here that is 2
plug in all the values in the above equation
E cell = 0.48 V - (0.059 / 2 ) (1.61)
E cell = 0.48 V - (0.0475 V)
= 0.43V
Get Answers For Free
Most questions answered within 1 hours.