Advertisements
Advertisements
Question
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]
Advertisements
Solution
Given: The given cell reaction is:
\[\ce{Mg_{(s)} | Mg^{2+} (0.1 M) || Cu^{2+} (0.01) | Cu_{(s)}}\]
Using the Nernst equation:
\[\ce{E_{cell} = E{^{\circ}_{cell}} - \frac{0.0591}{n} log \frac{[Mg^{2+}]}{[Cu^{2+}]}}\]
Where:
\[\ce{E^{\circ}_{cell}}\] = +2.71 V
n = 2 (number of electrons transferred)
Mg2+ = 0.1 M
Cu2+ = 0.01 M
\[\ce{E_{cell} = 2.71 - \frac{0.0591}{2} log \frac{0.1}{0.01}}\]
= 2.71 − 0.02955 log (10)
= 2.71 − 0.02955 × 1
= 2.71 − 0.02955
= 2.680 V
APPEARS IN
RELATED QUESTIONS
How much charge is required for the reduction of 1 mol of Zn2+ to Zn?
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.
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:
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 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: \[\ce{E^\circ Al^3+/Al = -1.66 V; E^\circ\phantom{.}Sn^2+/Sn = -0.14 V}\]
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)
Write the Nernst equation and emf of the following cell at 298 K:
Pt(s) | Br− (0.010 M) | Br2(l) || H+ (0.030 M) | H2(g) (1 bar) | Pt(s)
