The pressure of H₂ required to make the potential of H₂ - electrode zero in pure water at 298 K is:

In the electrochemical cell:

Zn|ZnSO₄(0.01 M) || CuSO₄(1.0M),Cu, the emf of this Daniel cell is E₁. When the concentration of ZnSO₄ is changed to 1.0 M and that of CuSO₄ is changed to 0.01 M, the emf changes to E₂. The relationship between E₁ and E₂ is : 
( Given, \(\frac{R T}{F}\)= 0.059)

1. E₁ = E₂

2. E₁ > E₂

3. E₁ < E₂

4. E₂ = 0 \(\neq\)E₁

The electrode potential of Cu electrode dipped in 0.025 M CuSO₄ solution at 298 K is:

(standard reduction potential of Cu = 0.34 V)

1. 0.047 V

2. 0.293 V

3. 0.35 V

4. 0.387 V

Consider the following cell reaction 

2Fe(s) + ${\mathrm{O}}_2$(g) + 4${\mathrm{H}}^{+}$(aq) $\to$ 2${\mathrm{Fe}}^{2+}$(aq) + 2${\mathrm{H}}_2\mathrm{O}$(l)

E° = 1.67 V, At [${\mathrm{Fe}}^{2+}$] = 10${}^{-3}$ M, ${\mathrm{P}}_{{\mathrm{O}}_2}$ = 0.1 atm and pH = 3, the cell potential at 25 °C is : 

1. 1.27 V

2. 1.77 V

3. 1.87 V

4. 1.57 V

The voltage of the cell given below increases with: 

Cell: Sn(s) + 2Ag⁺(aq) → Sn²⁺(aq) + 2Ag(s)

1. Increase in size of the silver rod.

2. Increase in the concentration of Sn²⁺ ions.

3. Increase in the concentration of Ag⁺ ions.

4. None of the above.

The electrode potential for the Mg electrode varies according to the equation:

\(E_{Mg^{2+}/Mg}\ = \ E_{Mg^{2+}/Mg}^{o} \ - \ \frac{0.059}{2}log\frac{1}{[Mg^{2+}]}\) 

The graph of E_{Mg²⁺ / Mg} vs log [Mg²⁺] among the following is:

1.		2.
3.		4.

By how much will the potential of half cell Cu²⁺| Cu change if the solution is diluted to 100 times at 298 K. It will -

The change in reduction potential of a hydrogen electrode when its solution initially at pH = 0 is neutralised to pH = 7, is a/an:

Calculate the EMF of the following concentration cell at 298 K:

Pt | H₂(1 atm) | H⁺(0.02 M) || H⁺(0.01 M) | H₂(1 atm) | Pt

1. - 0.017 V

2. 0.0295 V

3. 0.10 V

4. 0.059 V

Consider the given cell:

Pt(s) | H₂(g, 1 bar) | H⁺ (0.030 M) || Br⁻ (0.010 M) | Br₂(l) | Pt(s)

If the concentration of Br⁻ ions is doubled and the concentration of H⁺ ions is reduced to half of its initial value,
how will the emf of the cell change?

1. Two times
2. Four times
3. Eight times
4. Remains the same