The correct answer is 7
\(Q = \frac{[H^+]^2 }{[ Cu^{+2} ] pH_2} = \frac{10^{-6}}{C} pH_2 = 1\)
\(E = E^º_{cell} - \frac{0.06}{n} log Q\)
\(0.31 = 0.34 - \frac{0.06}{2} log \frac{10^{-6}}{C}\)
\(log \frac{10^{-6}}{C} = 1\)
C = 10-7 M
x = 7
The molar conductance of an infinitely dilute solution of ammonium chloride was found to be 185 S cm$^{-1}$ mol$^{-1}$ and the ionic conductance of hydroxyl and chloride ions are 170 and 70 S cm$^{-1}$ mol$^{-1}$, respectively. If molar conductance of 0.02 M solution of ammonium hydroxide is 85.5 S cm$^{-1}$ mol$^{-1}$, its degree of dissociation is given by x $\times$ 10$^{-1}$. The value of x is ______. (Nearest integer)
1 Faraday electricity was passed through Cu$^{2+}$ (1.5 M, 1 L)/Cu and 0.1 Faraday was passed through Ag$^+$ (0.2 M, 1 L) electrolytic cells. After this, the two cells were connected as shown below to make an electrochemical cell. The emf of the cell thus formed at 298 K is:
Given: $ E^\circ_{\text{Cu}^{2+}/\text{Cu}} = 0.34 \, \text{V} $ $ E^\circ_{\text{Ag}^+/ \text{Ag}} = 0.8 \, \text{V} $ $ \frac{2.303RT}{F} = 0.06 \, \text{V} $
Consider the following reaction:
\[ A + NaCl + H_2SO_4 \xrightarrow[\text{Little amount}]{} CrO_2Cl_2 + \text{Side products} \] \[ CrO_2Cl_2(\text{Vapour}) + NaOH \rightarrow B + NaCl + H_2O \] \[ B + H^+ \rightarrow C + H_2O \]
The number of terminal 'O' present in the compound 'C' is ________.
At steady state, the charge on the capacitor, as shown in the circuit below, is -----\( \mu C \). 
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.