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प्रश्न
How much charge is required for the reduction of 1 mol of Zn2+ to Zn?
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उत्तर
Zn2+ + 2e- → Zn
Number of electrons involved = 2
Charge required for the reduction of Zn2+ = 2F
We know
1F = 96,487 C
Thus,
2F = 2 × 96487 = 1,92,974 C
संबंधित प्रश्न
Write the Nernst equation and emf of the following cell at 298 K:
Mg(s) | Mg2+ (0.001 M) || Cu2+ (0.0001 M) | Cu(s)
Write the Nernst equation and explain the terms involved.
Calculate emf of the following cell at 298 K:
\[\ce{Mg_{(s)} | Mg^{2+} (0.1 M) || Cu^{2+} (0.01) | Cu_{(s)}}\]
[Given \[\ce{E^{\circ}_{cell}}\] = +2.71 V, 1 F = 96500 C mol–1]
Calculate the value of Ecell at 298 K for the following cell:
`(Al)/(Al^(3+)) (0.01M) || Sn^(2+) ((0.015 M))/(Sn)`
`E° _(Al^(3+))/(AI)= -1.66 " Volt and " E° _(Sn^(2+)) /(Sn) = -0.14` volt
For a general electrochemical reaction of the type:
\[\ce{{a}A + {b}B ⇔ {c}C + {d}D}\]
Nernst equation can be written as:
What is the pH of HCl solution when the hydrogen gas electrode shows a potential of −0.59 V at standard temperature and pressure?
Calculate ΔrG0 and log Kc for the following cell:
\[\ce{Ni(s) + 2Ag^+(aq) -> Ni^{2+}(aq) + 2Ag(s)}\]
Given that \[\ce{E^0_{cell}}\] = 1.05 V, 1F = 96,500 C mol–1.
Calculate the emf of the following cell at 298 K:
Fe(s) | Fe2+ (0.01 M) | | H+ (1 M) | H2(g) (1 bar) Pt(s)
Given \[\ce{E^0_{cell}}\] = 0.44 V.
Calculate the value of ΔG that can be obtained from the following cell at 298K.
`(Al)/(Al3+) (0.01M) || (Sn^(2+) (0.015M))/(Sn)`
`E^\circ (Al^(3+))/(Al) = -1.66 V; E^\circ (Sn^(2+))/(Sn) = -0.14 V`
Write the Nernst equation and emf of the following cell at 298 K:
Fe(s) | Fe2+ (0.001 M) || H+ (1 M) | H2(g) (1 bar) | Pt(s)
