Cell potential is the measure of the voltage difference between the anode and cathode.
\(\Delta G \, of \, H_2O(l)=-237.2 KJ/mol\)
\(\Delta G \, of \, CO_2(g)=-394.4 \, KJ/mol\)
\(\Delta G \, of \, pentane (g)=-8.2 kJ/mol\)
In pentane-oxygen fuel cell following reaction takes place
\(C_5H_{12}+10H_2O(l) \rightarrow5CO_2+32 H^+ \, +32e^-\)
\(\frac {8O_2+32H^+\, +32e^- \rightarrow 16H_2O(l)} {C_5H_{12}+8O_2 \rightarrow5CO^2+6H_2O(l), E^0=? }\)
\(\Delta G_{reaction}=\boldsymbol{\Sigma} \Delta G_{product}-\boldsymbol{\Sigma} \Delta G_{reactant}\)
\(=5 \times \Delta G_{(CO_2)}+6\Delta G_{(H_2O)}-[\Delta G_{(C_5H_{12})}+8 \times \Delta G_{O_2}]\)
\(=5 \times (-394.4)+6 \times (-237.2)-(-8.2 +0)\)
= - 1972- 1423.2+ 8.2
= -3 3 8 7 kJ/mol
\(\, \, \, \, \, \, \, \, =-3387 \times 10^3 J/mol\)
\(\Delta G=-nFE^0_{cell}\)
\(-3387 \times 10^3=-32 \times 96500 \times E^0_{cell}\)
\(\, \, \, \, \, E^0_{cell}= \frac {-3387 \times 10^3}{-32 \times 96500}=1.0968 V\)
Assertion (A): Cu cannot liberate \( H_2 \) on reaction with dilute mineral acids.
Reason (R): Cu has positive electrode potential.
The elements of the 3d transition series are given as: Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn. Answer the following:
Copper has an exceptionally positive \( E^\circ_{\text{M}^{2+}/\text{M}} \) value, why?
Match the Following
List-I (Use) | Item | Matches with | List-II (Substance) |
---|---|---|---|
A | Electrodes in batteries | II | Polyacetylene |
B | Welding of metals | III | Oxyacetylene |
C | Toys | I | Polypropylene |
An electrochemical cell is fueled by the combustion of butane at 1 bar and 298 K. Its cell potential is $ \frac{X}{F} \times 10^3 $ volts, where $ F $ is the Faraday constant. The value of $ X $ is ____.
Use: Standard Gibbs energies of formation at 298 K are:
$ \Delta_f G^\circ_{CO_2} = -394 \, \text{kJ mol}^{-1}; \quad \Delta_f G^\circ_{water} = -237 \, \text{kJ mol}^{-1}; \quad \Delta_f G^\circ_{butane} = -18 \, \text{kJ mol}^{-1} $
Consider the following electrochemical cell at standard condition. $$ \text{Au(s) | QH}_2\text{ | QH}_X(0.01 M) \, \text{| Ag(1M) | Ag(s) } \, E_{\text{cell}} = +0.4V $$ The couple QH/Q represents quinhydrone electrode, the half cell reaction is given below: $$ \text{QH}_2 \rightarrow \text{Q} + 2e^- + 2H^+ \, E^\circ_{\text{QH}/\text{Q}} = +0.7V $$
AB is a part of an electrical circuit (see figure). The potential difference \(V_A - V_B\), at the instant when current \(i = 2\) A and is increasing at a rate of 1 amp/second is:
An electrochemical cell is a device that is used to create electrical energy through the chemical reactions which are involved in it. The electrical energy supplied to electrochemical cells is used to smooth the chemical reactions. In the electrochemical cell, the involved devices have the ability to convert the chemical energy to electrical energy or vice-versa.