Original answer

var secondMax = function (){ 
    var arr = [20, 120, 111, 215, 54, 78]; // use int arrays
    var max = Math.max.apply(null, arr); // get the max of the array
    arr.splice(arr.indexOf(max), 1); // remove max from the array
    return Math.max.apply(null, arr); // get the 2nd max
};

demo

Update 1

As pointed out by davin the performance could be enhanced by not doing a splice but temporarily replacing the max value with -Infininty:

var secondMax = function (arr){ 
    var max = Math.max.apply(null, arr), // get the max of the array
        maxi = arr.indexOf(max);
    arr[maxi] = -Infinity; // replace max in the array with -infinity
    var secondMax = Math.max.apply(null, arr); // get the new max 
    arr[maxi] = max;
    return secondMax;
};

Anyway, IMHO the best algorithm is Jack's. 1 pass, with conversion to number. Mine is just short, using builtin methods and only wanted to provide it as an alternative, to show off all the different ways you can achieve the goal.

Update 2

Edge case with multiple values.

As comments pointed it out: this solution "does not work" if we have an array like [3, 3, 5, 5, 5, 4, 4]. On the other hand it would be also a matter of interpretation what we would consider "the 2nd largest element". In the example we have:

1. 3 elements with the largest value (5) at indices: 2,3,4
2. 2 elements with the second largest value (4) at indices: 5,6
3. 2 elements with the second smallest value (3) at indices: 1,2

The 2nd largest element could be interpreted as:

1. the 2nd (largest element) - 5 at index 3 - assuming that there is an order, and that we aim for a unique value
2. the (2nd largest) element - 4 at index 5 - assuming that there is an order, and that we aim for a unique value

Using ES6 Set and Array.from

const secondLargest = (arr) => Array.from([...new Set(arr)]).sort((a,b) => b-a)[1]

Above function removes duplicate elements using Set and returns the second largest element from the sorted array.