Calculate the standard potential, E∘, for this reaction from its △G∘ value. X(s)+Y3+(aq)⇒X3+(aq)+Y(s) △?∘=−28.0 kJ

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Calculate the standard potential, E∘, for this reaction from its △G∘ value.   X(s)+Y3+(aq)⇒X3+(aq)+Y(s)  △?∘=−28.0 kJ

jgardner33v4 · Accepted Answer

Step 1  E∘ and △G∘ are related as ÷  △G∘=−n F E∘  where n is no. of electrons transferred.  Step 2  Given, △G∘=−28.0kj  E∘=?  x+y3+⇒x3++y  Reaction at anade :  x⇒x3++3e−  ∴n=3  △G∘=−n F E∘  E∘=−△G∘n F  =−(−28.0)kJ3×96500C  =9.67×10−5kJc  =9.67×10−5×103Jc=0.0967V  Answer: 0.0967V

Samantha Brown · Accepted Answer

Step 1The reaction given isX(s)+Y4+(aq)⇒X4+(aq)+Y(s)In the above balanced reaction, 1 molecule of X (s) is being oxidised from 0 → 4+ oxidation state.Hence there are total 4 electrons required for the oxidation of X (s) from 0 → 4+.Hence in the reaction, total 4 electrons are transferred.Step 2The relationship between the Ecell0 and △G0 is given bynFEcell0=−△G0where n=number of electrons transferred in the reaction=4F=Faraday&#039;s constant=96500Cand △G0=91.0KJ=91000J (since 1KJ=1000J)Hence substituting the values we get4×96500×Ecell0=−91000⇒Ecell0=−0.23575V ≈

alenahelenash · Accepted Answer

Answer: 0.124 V Explanation: X(s)+Y4+(aq)⇒X4+(aq)+Y(s) To calculate standard Gibbs free energy, we use the equation: △G∘=−n F Ecello Where, n=number of electrons transferred=4 F=Faradays constant=96500C Ecello=standard cell potential=? Putting values in above equation, we get: −48.0×1000=−4×96500×Ecello Ecello=0.124 Thus the standard potential E∘, for this reaction is 0.124 V

Calculate the standard potential, E^{\circ}, for this reaction from its

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