Chemistry

Standard EMF of the cell = Standard Reduction Potential of the half cell on the right hand side (Cathode) - Standard Reduction Potential of the half cell on the left hand side (Anode)

For example, EMF of Daniell cell,

$E_{cell}=E_{Cu_{2+}∣Cu}−E_{Zn_{2+}∣Zn}$

If we were taking Oxidation Potentials into account,

Standard EMF of the cell = Standard Oxidation Potential of the half cell on the left hand side (Anode)-Standard Oxidation Potential of the half cell on the right hand side (Cathode). For example, EMF of Daniell cell,

$E_{cell}=E_{Zn∣Zn_{2+}}−E_{Cu∣Cu_{2+}}$

The Nernst Equation enables the determination of cell potential under non-standard conditions. It relates the measured cell potential to the reaction quotient and allows the accurate determination of equilibrium constants (including solubility constants).

Nernst equation is given as:

$E_{cell}=E_{cell}−nFRT logQ$

i.e. $E_{cell}=E_{cell}−n0.0591 logQ$ at $25_{o}C$

The equation above indicates that the electrical potential of a electrode depends upon the reaction quotient $Q$ of the reaction.

So if we are trying to determine reduction potential of $Cu$ electrode, we will have to consider $Cu_{2+}+2e_{−}→Cu_{(s)}$

We can then Nernst Equation as:

$E_{Cu_{2+}/Cu}=E_{Cu_{2+}/Cu}−20.0592 log([Cu_{2+}]1 )$ at $25_{o}C$

This way we can determine the $E_{Cu_{2+}/Cu}$ for Copper electrode at any temperature and concentration, if we know the standard reduction potential of Copper electrode i.e. $E_{Cu_{2+}/Cu}$

Nernst equation is given as:

$E_{cell}=E_{cell}−nFRT logQ$

i.e. $E_{cell}=E_{cell}−n0.0591 logQ$ at $25_{o}C$

The equation above indicates that the electrical potential of a cell depends upon the reaction quotient $Q$ of the reaction.

For a cell:

$Zn∣Zn_{2+}∣∣H_{+}∣H_{2}∣Pt$

We have a net chemical reaction of $Zn_{(s)}+2H_{+}→Zn_{2+}+H_{2(g)}$

If the concentrations of the ions are not 1.0 M, and the $H_{2}$ pressure is not 1.0 atm, then the cell potential may be calculated using the Nernst equation:

$E_{cell}=E_{cell}−20.0591 log([H_{+}]_{2}P(H_{2})[Zn_{2+}] )$

$Cd_{2+}+2e_{−}→Cd−E_{o}=−0.403V$

$Pb_{2+}+2e_{−}→Pb−E_{o}=−0.126V$ where, $[Cd_{2+}]=0.02M,[Pb_{2+}]=0.2M$

Solution:

The first step is to determine the cell reaction and total cell potential.

In order for the cell to be galvanic, reactions need to be spontaneous i.e. $E_{cell}>0$.

Since Cadmium is having lesser reduction potential amongst the two, Cadmium must undergo oxidation. Hence reactions involved will be:

$Cd→Cd_{2+}+2e_{−},Pb_{2+}+2e_{−}→Pb$

Hence, overall reaction will be:

$Pb_{2+}+Cd→Pb+Cd_{2+}$

$E_{cell}=+0.403−0.126=0.277V$

Now, from Nernst Equation we have,

$E_{cell}=E_{cell}−n0.0591 logQ$ at $25_{o}C$

Here, we can write Nernst Equation as,

$E_{cell}=0.277−20.059 log(0.20.02 )$

i.e. $E_{cell}=0.300V$

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