If `E_(cell)^(ɵ)` for a given reaction is negative, which gives the correct relationships for the values of `DeltaG^(ɵ)` and `K_(eq)`?A. `DeltaG^(@) gt 0 : K_(eq) lt 1`B. `DeltaG^(@) gt 0, K_(eq) gt 1`C. `DeltaG^(@) lt 0, K_(eq) gt 1`D. `DeltaG^(@) lt 0, K_(eq) lt 1`

If `E_(cell)^(ɵ)` for a given reaction is negative, which gives the c

1.	If `E_(cell)^(ɵ)` for a given reaction is negative, which gives the correct relationships for the values of `DeltaG^(ɵ)` and `K_(eq)`?A. `DeltaG^(@) gt 0 : K_(eq) lt 1`B. `DeltaG^(@) gt 0, K_(eq) gt 1`C. `DeltaG^(@) lt 0, K_(eq) gt 1`D. `DeltaG^(@) lt 0, K_(eq) lt 1`
Answer» Correct Answer - A `DeltaG^(@)=-nFE^(@)` `E^(@)` is negative So `DeltaG^(@)=+ve` `DeltaG^(@)=-2.303RT" log "K` `thereforeK_(eq) lt1`

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If `E_(cell)^(ɵ)` for a given reaction is negative, which gives the correct relationships for the values of `DeltaG^(ɵ)` and `K_(eq)`?A. `DeltaG^(@) gt 0 : K_(eq) lt 1`B. `DeltaG^(@) gt 0, K_(eq) gt 1`C. `DeltaG^(@) lt 0, K_(eq) gt 1`D. `DeltaG^(@) lt 0, K_(eq) lt 1`

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