Explanation The amount of electrons needed for reduction and stoichiometric relation between amount of electrons and metal produced from the reduction half-reaction. The required supplied electrons can be calculated by using following the equation is, n = It F Where, T= time N=number of electrons F= faraday I= current Now, write the reduction reaction of silver ion is as follows, The amount of electrons n(e - ) supplied from the following equation n = It F time (t) = 13 .7 h × 3600 s 1 h (1h = 3600 s) Current (I)=350 mA =350× 10 -3 A (1 mA = 10 -3 A) Faradays constant (F) = 9 .6485×10 4 C ⋅ (mol .e - ) -1 . The above equation becomes, n = I t F = (350× 10 -3 A) ( 37 .1 h × 3600 s 1 h ) 9 .6485×10 4 C ⋅ (mol .e - ) -1 = (350× 10 -3 A) ( 13 .7 h × 3600 s 1 h ) 9 .6485×10 4 A .s ⋅ (mol .e - ) -1 ( ∵ 1C = 1 A .s ) = 350× 10 -3 × 13

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM

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ISBN: 9781319403959

Author: ATKINS

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Chapter 6, Problem 6O.12E

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The oxidation number of manganese in the given quantity of manganese nitrate has to be identified.

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Chapter 6, Problem 6O.11E

Chapter 6, Problem 6O.13E

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ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM

Ch. 6 - Prob. 6A.1AST Ch. 6 - Prob. 6A.1BST Ch. 6 - Prob. 6A.2AST Ch. 6 - Prob. 6A.2BST Ch. 6 - Prob. 6A.3AST Ch. 6 - Prob. 6A.3BST Ch. 6 - Prob. 6A.1E Ch. 6 - Prob. 6A.2E Ch. 6 - Prob. 6A.3E Ch. 6 - Prob. 6A.4E

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The oxidation number of manganese in the given quantity of manganese nitrate has to be identified.

Solution Summary: The author explains that the oxidation number of manganese in the given quantity of nitrate has to be identified. The amount of electrons needed for reduction can be calculated by using the following equation.

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Interpretation: The oxidation number of manganese in the given quantity of manganese nitrate has to be identified.

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Chapter 6 Solutions

Interpretation:

The oxidation number of manganese in the given quantity of manganese nitrate has to be identified.