First slide

Electrochemical cells

Question

A 60 ml sample solution of KI was electrolysed for 579 sec using a 5 amp constant current. The produced, completely reacted with 0.20 M sodium thiosulphate solution and unreacted KI solution required 30 ml of 0.04 M acidic solution of KMnO₄. If the original molarity of KI solution can be expressed in scientific notation as Y × 10^-1. Calculate the value of Y.

Moderate

Solution

$\begin{array}{l} n_{F} = \frac{579 \times 5}{96500} = 0.03 \\ 2 I^{-} \to I_{2} + 2 e^{-} \\ n_{F} = 2 n_{I_{2}} \\ n_{I_{2}} = 00.015 mol \Rightarrow n_{KI} = 0.03 = 30 \times 10^{- 3} \\ e_{I^{-}} = e_{{KMNO}_{4}} \\ n_{I^{-}} = \frac{30}{1000} \times .04 \times 5 = 6 \times 10^{- 3} \\ [KI] = \frac{36 \times 10^{- 3} \times 10^{3}}{60} = 0.6 \\ [KI] = 6 \times 10^{- 1} M \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App