Advertisements
Advertisements
प्रश्न
How much charge is required for the reduction of 1 mol of Zn2+ to Zn?
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:
\[\ce{Mg_{(s)} | Mg^{2+} (0.001 M) || Cu^{2+} (0.0001 M) | Cu_{(s)}}\]
Write the Nemst equation and explain the terms involved.
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
Complete the following statement by selecting the correct alternative from the choices given:
For a spontaneous reaction ΔG° and E° cell will be respectively:
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 the emf of the following cell at 298 K.
\[\ce{Cu/Cu^{2+}_{(0.025 M)}//Ag^+_{(0.005 M)}/Ag}\]
Given `"E"_("Cu"^(2+)//"Cu")^circ` = 0.34 V, `"E"_("Ag"^+//"Ag")^circ` = 0.80 V
1 Faraday = 96500 C mol -1
Calculate the value of \[\ce{E^\circ}\]cell, E cell and ΔG that can be obtained from the following cell at 298 K.
\[\ce{Al/Al^3+ _{(0.01 M)} // Sn^{2+} _{(0.015 M)}/Sn}\]
Given: \[\ce{E^\circ Al^3+/Al = -1.66 V; E^\circ\phantom{.}Sn^2+/Sn = -0.14 V}\]
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:
\[\ce{Fe_{(s)} | Fe^{2+} (0.001 M) || H^+ (1 M) | H2_{(g)} (1 bar) | Pt_{(s)}}\]
