Calculate the EMF of the cell: $Zn|Zn^{2+}(0.1 M)||Cu^{2+}(0.01 M)|Cu$
Calculate the EMF of the cell: $Zn|Zn^{2+}(0.1 M)||Cu^{2+}(0.01 M)|Cu$
Given: $E°_{Zn^{2+}/Zn} = -0.76$ V, $E°_{Cu^{2+}/Cu} = +0.34$ V
1 Answer
VAVidaara Admin
✓ Vidaara Team
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· 3mo ago
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$$E°_{cell} = E°_{cathode} - E°_{anode} = 0.34 - (-0.76) = 1.10 V$$
Using Nernst equation ($n = 2$, $T = 298$ K):
$$E = E° - \frac{0.0591}{n}\log Q = 1.10 - \frac{0.0591}{2}\log\frac{[Zn^{2+}]}{[Cu^{2+}]}$$
$$= 1.10 - 0.02955\times\log\frac{0.1}{0.01} = 1.10 - 0.02955\times1$$
$$E = 1.07 V$$
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Discussion (3)
PP
Is there a faster shortcut for this in the actual exam? Time is tight.
E
Saved me before my mock test. Much clearer than my coaching notes.
A
Saved me before my mock test. Much clearer than my coaching notes.