JEE Main Chemistry question
where for is V, and for is 0.150 V. The emf of the cell is
Answer: (D)
Solution
E_{cell} = E^\circ_{cathode}-E^\circ_{anode} = E^\circ_{Au^{3+}/Au}-E^\circ_{Ni^{2+}/Ni} = 0.150-(-0.250) = +0.4\ V.
where for is V, and for is 0.150 V. The emf of the cell is
E_{cell} = E^\circ_{cathode}-E^\circ_{anode} = E^\circ_{Au^{3+}/Au}-E^\circ_{Ni^{2+}/Ni} = 0.150-(-0.250) = +0.4\ V.