1896 United States presidential election in Vermont

The 1896 United States presidential election in Vermont took place on November 3, 1896 as part of the 1896 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.

1896 United States presidential election in Vermont

November 3, 1896

Nominee William McKinley William Jennings Bryan Party Republican Democratic Home state Ohio Nebraska Running mate Garret Hobart Arthur Sewall Electoral vote 4 0 Popular vote 51,127 10,640 Percentage 80.08% 16.66%

County Results McKinley   60-70%   70-80%   80-90%

President before election Grover Cleveland Democratic Elected President William McKinley Republican

Elections in Vermont
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Municipal offices Mayoral elections City of Burlington 1865 1981 1983 1985 1987 1989 1991 1993 1995 1997 2000 2003 2006 2009 2012 2015 2018 2021 City of Rutland 2017 2019

Vermont overwhelmingly voted for the Republican nominee, former Governor of Ohio William McKinley, over the Democratic nominee, former U.S. Representative from Nebraska William Jennings Bryan. With 80.08% of the popular vote, Vermont would be McKinley's strongest victory in terms of percentage in the popular vote as well as the best performance of any candidate Green Mountain State to date.[1] McKinley won Vermont by a landslide margin of 63.42%, which is the second-best performance by the Republican Party in any presidential election behind only Barry Goldwater in Mississippi in 1964 – ironically the first election when the Green Mountain State deserted the GOP.

Bryan, running on a platform of free silver, appealed strongly to Western miners and farmers in the 1896 election, but had practically no appeal in the Northeastern state of Vermont. In fact, with only 16.66% of the popular vote Bryan's performance in Vermont was the second weakest of any Democratic presidential candidate to date only after John W. Davis' 15.67% in 1924.