Question

33. A hydrogen gas electrode is made by dipping
platinum wire in a solution of HCl of pH = 10 and
by passing hydrogen gas around the platinum
wire at one atm pressure. The oxidation potential
of the electrode would be
​

Answer 1

Answer:

here is your answer

Explanation:

The solution of HCl has pH=10Hence, [H + ]=10 −pH =10 −10 MThe pressure of hydrogen gas is one atm. The oxidation potential of electrode would be E cell = n0.059 log [H + ](P H 2 ) 0.5 = 1

0.059

0.059

0.059 log

0.059 log 10

0.059 log 10 −10

0.059 log 10 −10

0.059 log 10 −10 1

0.059 log 10 −10 1

0.059 log 10 −10 1

0.059 log 10 −10 1 =+0.59V.

0.059 log 10 −10 1 =+0.59V.Hence, the oxidation potential of hydrogen electrode is +0.59 V.

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Answer 2

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• The solution of HCl has pH=10

Hence,

• $[H + ]=10 </li></ul><p>−pH</p><p> =10 </p><p>−10</p><p> M$
  • The pressure of hydrogen gas is one atm.
  • The oxidation potential of electrode would be
  • $E </li></ul><p>cell</p><p> </p><p> = </p><p>n</p><p>0.059</p><p> </p><p> log \: </p><p>[H </p><p>+</p><p> ] \: </p><p>(P </p><p>H </p><p>2</p><p> </p><p> </p><p> </p><p> ) </p><p>0.5 \\ </p><p> </p><p> </p><p> = </p><p>1</p><p>0.059</p><p> </p><p> \: log \: </p><p> \: 10 </p><p>−10</p><p> </p><p>1</p><p> \\ </p><p> </p><p></p><p>=+0.59V$

    .

    • Hence, the oxidation potential of hydrogen electrode is +0.59 V.

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