Chemistry, 06.09.2019 21:10, sheram2010

Determine whether the following are spontaneous or non-spontaneous under standard conditions:
ca(s) + cd2+(aq) ® ca2+(aq) + cd(s)
2 br-(aq) + sn2+(aq) ® sn(s) + br2(l)

Answers: 2

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Answers

answered: erikloza12pdidtx

For 1: The above reaction is non-spontaneous.

For 2: The above reaction is non-spontaneous.

Explanation:

For 1:

The given chemical reaction: follows:

$Cd^{2+}(aq.)+Ca(s)\rightarrow Ca^{2+}(aq.)+Cd(s)$

Here, cadmium is getting reduced because it is gaining electrons and calcium is getting oxidized because it is loosing electrons.

We know that:

$E^o_{(Cd^{2+}/Cd)}=-0.40V\\E^o_{(Ca^{2+}/Ca)}=-2.87V$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=-0.40-(-2.87)=2.47V$

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$

As, the standard electrode potential of the cell is coming out to be positive for the above cell. Thus, the standard Gibbs free energy change of the reaction will become negative making the reaction spontaneous.

Hence, the above reaction is spontaneous.

For 2:

The given chemical reaction: follows:

$2Br^-(aq.)+Sn^{2+}(aq.)\rightarrow Br_2(l)+Sn(s)$

Here, tin is getting reduced because it is gaining electrons and bromine is getting oxidized because it is loosing electrons.

We know that:

$E^o_{(Sn^{2+}/Sn)}=-0.14V\\E^o_{(Br_2/2Br^-)}=-1.07V$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=-0.14-(1.07)=-1.21V$

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nfE^o_{cell}$

As, the standard electrode potential of the cell is coming out to be negative for the above cell. Thus, the standard Gibbs free energy change of the reaction will become positive making the reaction non-spontaneous.

Hence, the above reaction is non-spontaneous.

answered: Guest

answer; capital earns interest; entrepreneurship receives a profit or incurs a loss;

answered: Guest

Answerit curves up and levels out.explanationwe assume that the container contains absolutely no ammonia to begin with as it states that "nitrogen gas and hydrogen gas" react to "produce ammonia."at dynamic equilibrium, the product and reactant concentrations are equal.this means that, starting from a concentration of zero, [nh₃] must curve "up" from zero and then level out to a constant concentration.