\(\frac{RT}{nF}InK_{eq}\) The above equation is called :

Gibb’s equation
Gibb’s–Helmholtz equation
Nernst equation
Van der Waal’s equation

Solution

Following cell has EMF 0.7995V. Pt | H₂(1 atm) | HNO₃(1M) || AgNO₃(1M) | Ag
If we add enough KCl to the Ag cell so that the final Cl– is 1M. Now the measured emf of the cell is 0.222V. The Ksp of AgCl would be –

1× 10^-9.8
1× 10^-19.6
2× 10^-10
2.64 × 10^-14

Solution

A hydrogen electrode is immersed in a solution with pH =0 (HCl). By how much will the potential (reduction)change if an equivalent amount of NaOH is added to the solution. (Take PH₂= 1 atm), T= 298 K.

increase by 0.41 V
increase by 59 mV
decrease by 0.41 V
decrease by 59 mV

Solution

The standard cell potential of Zn|Zn²⁺(aq)||Cu²⁺(aq)|Cu cell is 1.10 V. The maximum work obtained by this cell will be(a)(b)(c)(d)

106.15 kJ
–212.30 kJ
–318.45 kJ
– 424.60 kJ

Solution

Adding powered lead and iron to a solution that is 1.0 M in both Pb²⁺ and Fe²⁺ ions, would result in a reaction, in which:

More iron and Pb²⁺ions are formed
More lead and Fe²⁺ions are formed
Concentration of both Pb²⁺and Fe²⁺ ion increases
There is no net change

Solution

Fe is above Pb in the electrochemical series i.e. E^°_oxide of Fe (+ 0.44 V) is higher than E^°_oxide of Pb (+ 0.129 V).Hence, on addition of powdered Fe and Pb to a solution of Fe⁺⁺ ion and Pb⁺⁺ ion, the following reaction will take place :
Fe(s) + Pb⁺⁺(aq) ⟶ Fe⁺⁺(aq) + Pb(s)
So, more Pb and Fe⁺⁺ ions will be formed.

For the galvanic cell Zn | Zn²⁺(0.1M) || Cu²⁺(1.0M)|Cu the cell potential increase if:

[Zn²⁺] is increased
[Cu²⁺] is increased
[Cu²⁺] is decreased
surface area of anode is increased

Solution

1.006 V
0.503 V
0.235 V
–0.235 V

Solution

The standard emf of a cell, involving one electron change is found to be 0.591 V at 25°C. The equilibrium constant of the reaction is (F = 96500 C mol^-1)

1.0 × 10¹
1.0 × 10⁵
1.0 × 10¹⁰
1.0 ×10³⁰

Solution

If a salt bridge is removed between the two half cells, the voltage

Drops to zero
Does not change
Increase gradually
Increases rapidly

Solution

On removing the salt bridge between the two half cells the circuit is broken.Hence, emf becomes zero.

Solution

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