The reaction for the deposition of zinc is as follows:
Zn2+ + 2e- → Zn
Using the formula for electrolysis:
W = \( \frac{Z \times i \times t}{F} \)
where
Calculating the mass of zinc:
W = \( \frac{65.4}{2 \times 96500} \times 0.015 \times 15 \times 60 \)
W = 45.75 $\times$ 10-4 g
Since the answer can be approximated, we also consider 46 $\times$ 10-4 g.
So, the correct answer is: 45.75 or 46
For the given cell: \[ {Fe}^{2+}(aq) + {Ag}^+(aq) \to {Fe}^{3+}(aq) + {Ag}(s) \] The standard cell potential of the above reaction is given. The standard reduction potentials are given as: \[ {Ag}^+ + e^- \to {Ag} \quad E^\circ = x \, {V} \] \[ {Fe}^{2+} + 2e^- \to {Fe} \quad E^\circ = y \, {V} \] \[ {Fe}^{3+} + 3e^- \to {Fe} \quad E^\circ = z \, {V} \] The correct answer is:
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32