A voltaic cell is constructed in which the anode is a Fe2+|Fe3+ half cell and the cathode is a Br-|Br2 half cell. The half-cell compartments are connected by a salt bridge.

(Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank.)

The anode reaction is:

(aq)(s)(l)(g) + (aq)(s)(l)(g) > (aq)(s)(l)(g) + (aq)(s)(l)(g)

The cathode reaction is:

(aq)(s)(l)(g) + (aq)(s)(l)(g) > (aq)(s)(l)(g) + (aq)(s)(l)(g)

The net cell reaction is:

(aq)(s)(l)(g) + (aq)(s)(l)(g) > (aq)(s)(l)(g) + (aq)(s)(l)(g)

In the external circuit, electrons migrate (from/to) the Fe2+|Fe3+ electrode (from/to) the Br-|Br2 electrode.

In the salt bridge, anions migrate (from/to) the Br-|Br2 compartment (from/to) the Fe2+|Fe3+ compartment.

#2
A voltaic cell is constructed in which the following cell reaction occurs. The half-cell compartments are connected by a salt bridge.

I2(s) + Pb(s) --->2I-(aq) + Pb2+(aq)

The anode reaction is:

(aq)(s)(l)(g) + (aq)(s)(l)(g) > (aq)(s)(l)(g) + (aq)(s)(l)(g)

The cathode reaction is:

(aq)(s)(l)(g) + (aq)(s)(l)(g)
> (aq)(s)(l)(g) + (aq)(s)(l)(g)

In the external circuit, electrons migrate (from/to) the Pb|Pb2+ electrode (from/to) the I-|I2 electrode.

In the salt bridge, anions migrate (from/to) the Pb|Pb2+ compartment (from/to) the I-|I2 compartment.

b) mutualism

hope this : )

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol