JEE MAIN - Chemistry (2002 - No. 33)
EMF of a cell in terms of reduction potential of its left and right electrodes is :
E = Eleft - Eright
E = Eleft + Eright
E = Eright - Eleft
E = -(Eright + Eleft)
Selitys
$${E_{cell}} = \,\,$$ Reduction potential of cathode (right)
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$-$$ Reduction potential of anode (left)
$$ = {E_{right}} - {E_{left}}.$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$-$$ Reduction potential of anode (left)
$$ = {E_{right}} - {E_{left}}.$$