Nernst equation shows a deviation from the practial value of emf of a cell if the electrolyte has a very high concentration

True

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False

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The correct option is A True

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Assertion: The Nernst equation gives concentration dependence of EMF of the cell.

Reason: In a Daniell cell current flows from cathode to anode.

Q. Assertion :Increase in the concentration of copper half cell in daniel cell increases the emf of the cell. Reason: According to Nernst equation.

E_{c e l l} = E {_{c e l l}}^{0} + \frac{0.059}{2} l o g \frac{[C u^{2 +}]}{[Z n^{2 +}]}

Q. Electra has mastered the basics of Nernst equation. She is now studying equilibrium constants relation with Nernst equation.

Electra was performing experiments with Nernst with a Daniel Cell. She noticed that when the reaction reaches equilibrium, the cell potential goes to 0.

She asked Nernst why this happened. His reply was

Figure 3.33 shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents up to a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

(a) What is the value ε ?

(b) What purpose does the high resistance of 600 kΩ have?

(d) Is the balance point affected by the internal resistance of the driver cell?

(e) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V?

(f ) Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit?

Q. Figure shows a potentiometer with a cell of 2.0 V and internal resistance

0.40 Ω

maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents up to a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 k

Ω

is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf and the balance point found similarly, turns out to be at 82.3 cm length of the wire.(a) What is the value

ε

?(b) What purpose does the high resistance of 600 k

Ω

have?(c) Is the balance point affected by this high resistance? (d) Is the balance point affected by the internal resistance of the driver cell?(e) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V?(f) Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit?

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