Knowee
Questions
Features
Study Tools

The EMF of the cell : Ni(s) | Ni2+ (1.0M) || Au 3+ (1.0M) | Au (s) is[ given EoNi2+/Ni -0.25V; E° Au3+ / Au = +1.50V ]

Question

The EMF of the cell : Ni(s) | Ni2+ (1.0M) || Au 3+ (1.0M) | Au (s) is[ given EoNi2+/Ni -0.25V; E° Au3+ / Au = +1.50V ]

🧐 Not the exact question you are looking for?Go ask a question

Solution

To find the EMF of the cell, we can use the Nernst equation:

Ecell = E°cell - (0.0592/n) * log(Q)

First, let's determine the half-reactions occurring at each electrode:

At the anode (left side): Ni(s) -> Ni2+(aq) + 2e-

At the cathode (right side): Au3+(aq) + 3e- -> Au(s)

Now, we can calculate the cell potential using the given standard reduction potentials:

E°cell = E°cathode - E°anode E°cell = E°Au3+/Au - E°Ni2+/Ni E°cell = +1.50V - (-0.25V) E°cell = +1.75V

Next, we need to calculate the reaction quotient, Q, using the concentrations of the species involved:

Q = [Ni2+]/[Au3+]

Since both concentrations are given as 1.0M, Q = 1.0/1.0 = 1.0

Now, we can substitute the values into the Nernst equation:

Ecell = E°cell - (0.0592/n) * log(Q)

Since the number of electrons transferred in the balanced equation is 2, n = 2:

Ecell = 1.75V - (0.0592/2) * log(1.0) Ecell = 1.75V - (0.0296) * log(1.0) Ecell = 1.75V - 0 Ecell = 1.75V

Therefore, the EMF of the cell is 1.75V.

This problem has been solved

Similar Questions

Find the emf of the cell in which the following reaction takes place at 298 K Ni(s) + 2 Ag+ (0.001 M) → Ni2+ (0.001 M) + 2 Ag (s)

At 298K, the EMF of the cell: Mg(s) |Mg2+(Q)||Ag+(0.01)|Ag(s)is 3.022V. Calculate the value ‘Q’.(Given: EoMg2+/Mg = -2.37V and EoAg+/Ag = 0.80V)

The standard EMF for the given cell reaction Zn+Cu2+⇒Cu+Zn2+ is1.10 V at 25∘C. The EMF for the cell reaction, when 0.1MCu2+ and 0.1M Zn2+ solutions are used, at 25∘C is1.10 V0.110 V−1.10 V−0.110 V

The standard electrode potentials for the reactionsAg+ (a) + e– ⎯⎯→ Ag(s) Sn2+ (a) + 2e– ⎯⎯→ Sn (s)at 25 °C are 0.80 volt and –0.14 volt, respectively. The standard emf of the cell formed by thesehalf-cells is

The emf of the cell,               Tl(s) | Tl+(aq) (0.0001 M) || Cu2+(aq) (0.01M) | Cu(s) is 0.83 V                The emf of this cell will be increased by :–Increasing the concentration of Cu2+ ionsDecreasing the concentration of Tl+Increasing the concentration of both(A) & (B) both

1/2

Upgrade your grade with Knowee

Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.