Nernst Equation

We are given with following cell reaction:

2H⁺ + 2e^–   $\stackrel{}{\to }$ H₂

_{$P_{H_2}$ =2 atm}

[H⁺] = 1 M

$\left(\frac{2.303RT}{F}=0.06\right)$

If E_cell for reaction is given by –x * 10^–3 V, find out x.

The cell potential for the given cell at 298 K

$Pt|H_2\left(g,1bar\right)\left|H^{+}\left(aq\right)\right|\left|C{u}^{2+}\left(aq\right)\right|Cu\left(s\right)is0.31V.$ The pH of the acidic solution is found to be 3, whereas the concentration of Cu²⁺ is 10^-x M. The value of x is________.

(Given : $E_{C{u}^{2+}/Cu}^{\Theta }=0.34Vand\frac{2.303RT}{F}=0.06V)$  

The cell potential for the given cell at 298 K

$Pt|H_2\left(g,1bar\right)\left|H^{+}\left(aq\right)\right|\left|C{u}^{2+}\left(aq\right)\right|Cu\left(s\right)is0.31V.$ The pH of the acidic solution is found to be 3, whereas the concentration of Cu²⁺ is 10^-x M. The value of x is________.

(Given :  $E_{C{u}^{2+}/Cu}^{\Theta }=0.34Vand\frac{2.303RT}{F}=0.06V)$  

For the cell Cu(s) | Cu²⁺ (aq)(0.1M) Ag⁺ (aq)(0.01M) | Ag(s)
The cell potential E₁ = 0.3095V
For the cell Cu(s) | Cu²⁺ (aq)(0.01M) Ag⁺ (aq)(0.001M) | Ag(s)
The cell potential = _____ *10⁻² V. (Round off to the Nearest Integer).
[Use : (2.303RT/F) = 0.059 ]

Find the emf of the cell in which the following reaction takes place at 298 K
Ni(s) + 2Ag⁺(0.001M) → Ni²⁺(0.001M) + 2Ag(s)
(Given that E°cell = 1.05 V, 2.303RT/F = 0.059 at 298 K)

A reaction of 0.1 mole of Benzylamine with bromomethane gave 23g of Benzyl trimethyl ammonium bromide. The number of moles of bromomethane consumed in this reaction are n x 10?¹, when n=_____. (Round off to the nearest integer).
[Given: Atomic masses: C=12.0u, H:1.0u, N:14.0u, Br:80.0u]

Kindy consider the following

One of the by-products formed during the recovery of NH₃ from solvay process is:

Kraft temperature is the temperature:

What would be the electrode potential for the given half cell reaction at $\mathrm{p}\mathrm{H}=5$ $$ ?